3
$\begingroup$

I'd like to compute the limit of

$$\mathop {\lim }\limits_{x \to +\infty } \frac{{\Gamma \left( {\frac{1}{2} - \frac{x}{2}} \right)\Gamma \left( {1 + \frac{x}{2}} \right)}}{{\Gamma \left( {\frac{1}{2} + \frac{x}{2}} \right)\Gamma \left( {\frac{x}{2}} \right)}}$$

Because of the physical problem involved, I can assume that x is even, i.e. x=2n (n positive non null integer) but... does it make sense to "approach infinity on even numbers"? Or this apparently oscillating function between +inf and –inf can somehow balance to zero? I’m little bit confused... Thanks in advance for any you can provide. Regards

$\endgroup$
  • $\begingroup$ It certainly makes sense to ask for $$\lim_{n\to\infty}{\Gamma((1/2)-n)\Gamma(1+n)\over\Gamma((1/2)+n)\Gamma(n)}$$ Whether it exists, or what it comes to, I don't know. What do you mean by "Oscillating between $\infty$ and $-\infty$"? $\endgroup$ – Gerry Myerson Apr 24 '14 at 23:49
  • $\begingroup$ I just tried a plot, it goes to +-infinity on odd numbers (1, 3 ...) passing the 0 on even numbers (0, 2...). Not sure it helps, maybe it's misleading me to believe that the limit is not indeterminate while it actually is... Any idea? Is there a way to split limits on integer between odd and even numbers? or to relate the limit to some series representation? Thanks $\endgroup$ – Ste Apr 25 '14 at 14:17
  • $\begingroup$ You say in the body of the question that you can assume $x$ is an even integer. So let $x=2n$, and you get the expression I gave in my comment. As long as the limit is understood as the limit of a sequence ($n\to\infty$ through the integers) and not as the limit of a function ($n\to\infty$ through the reals), there is no difficulty here, and there's nothing wrong with saying the limit is zero. $\endgroup$ – Gerry Myerson Apr 26 '14 at 4:32
  • $\begingroup$ Yep, but at least to reduce my question to prove that on odd numbers the +-inf balance to zero would simplify my physic result: or is it going as, say $(-1)^n$, and there's no hope? I start thinking to give up the search of the limit on reals :-/ Thanks $\endgroup$ – Ste Apr 26 '14 at 8:47
  • $\begingroup$ I don't know what "balance to zero" means. If you have the sequence $1,0,-2,0,3,0,-4,0,5,\dots$ then your sequence doesn't converge, but the subsequence $0,0,0,0,\dots$ of even terms converges to zero, and that's about all you can do. What I really don't understand is why you worry about the odd terms and the reals when you say you can assume $x$ is a positive even integer. $\endgroup$ – Gerry Myerson Apr 26 '14 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.